Question

Value of ${\int }_0^{\pi }|\sin x-\cos x|dx$

• A. $0$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. $1/\sqrt{2}$

Answer 1

Answer: (C)

$2\sqrt{2}$

${\int }_0^{\pi }\left|\sin x-\cos x\right|dx={\int }_0^{\frac{\pi }{4}}|\sin x-\cos x|dx+{\int }_{\frac{\pi }{4}}^{\pi }|\sin x-\cos x|dx$

$=-{\int }_0^{\frac{\pi }{4}}(\sin x-\cos x)dx+{\int }_{\frac{\pi }{4}}^{\pi }(\sin x-\cos x)dx$

$=-{[-\cos x-\sin x]}_0^{\frac{\pi }{4}}+{[-\cos x-\sin x]}_{\frac{\pi }{4}}^{\pi }$

$=-[\frac{-1}{\sqrt{2}}\frac{-1}{\sqrt{2}}+1-0]+[+1-0+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}]$

$=+\sqrt{2}-1+1+\sqrt{2}=2\sqrt{2}$